Six Sigma and Beyond: Statistics and Probability, Volume III

OVERVIEW

Problems of maximizing or minimizing a function of several variables when those variables satisfy some restriction equations can become quite difficult to solve. There is a technique, called the method of Lagrange multipliers, which often simplifies the calculations. It is named after the eighteenth-century French mathematician who introduced it.

For the purpose of illustrating the technique, we solve a simple problem involving only two variables. We wish to find where the function f(x,y) = x ² + 2xy assumes its minimum value if the variables are subject to the restriction equation x ² y = 27. This problem can be solved by solving for y in terms of x in the restriction equation and substituting it into f(x,y) to reduce f to a function of a single variable.

In the Lagrange multiplier method a new function is introduced in the following manner. First, we rewrite the restriction equation so that it assumes the form g(x,y) = 0. Hence, we write

g(x,y) = x ² y - 27 = 0

The new function then is the function defined by

The parameter » (lambda) is the Lagrange multiplier here. It always multiplies the restriction function after the restriction equation has been expressed in the form g(x,y) = 0. We now treat F(x,y) as though it were a function of x and y without any restriction on those variables. We therefore proceed to find where the function F(x,y) assumes its minimum value in the manner of the preceding sections. That is, we calculate F _x and F _y , set them equal to zero, find the critical points, and check to see which of those points, if any, yields the minimum. In this problem, we obtain

Since we have the restriction equation

x ² y - 27 = 0

in addition to the preceding two equations, we have three equations in the three unknowns: x, y, and lambda ( » ). We proceed to solve them. Since we are not interested in the value of » , we try to eliminate it first. Solving for » in the second of the two partial derivative equations, we obtain

This is substituted into the first of those two equations to give

2x + 2y - 4y = 0

which reduces to

x = y

The preceding three equations in three unknowns are now reduced to the following two equations in two unknowns:

x = y

x ² y - 27 = 0

The solution of this pair of equations is x = 3 and y = 3. As previously, it is easily shown that this pair of numbers minimizes f(x,y) = x ² + 2xy, subject to the restriction x ² y = 27.

The reasoning behind the Lagrange multiplier technique is the following one. Suppose that we have found a point (x ,y ) that minimizes f(x,y) subject to the restriction g(x,y) = 0. Then this point must also minimize the function F(x,y) regardless of the value of » because as long as the restriction g(x,y) = 0 is satisfied, the term » g(x,y) has the value zero and, hence, minimizing f(x,y) minimizes F(x,y). Conversely, any point (x ,y ) that minimizes F(x,y) and that also satisfies the restriction equation must minimize f(x,y) subject to this restriction because once more the term » g(x,y) has the value zero for any such point. Therefore if F(x ,y ) is a minimum, f(x ,y ) must be a minimum. Although this shows that minimizing F is equivalent to minimizing f, it does not prove that we may treat F as a function of x and y without any restrictions on those variables. A proof of this fact is rather involved and, therefore, will not be given here.

The same type of reasoning applies to problems involving functions of several variables and with several restrictions. We demonstrate the technique on the following problem.

Given the function f(x,y,z) = xyz, subject to the restriction xy + 2xz + 2yz = 24, find where it is maximized. We first write the restriction in the form of g(x,y,z) = xy + 2xz + 2yz - 24 = 0. Then we write

F ( x , y , z ) = f ( x , y , z ) - » g ( x , y , z )

= xyz - » [ xy + 2 xz + 2 yz - 24]

Next, we calculate the three partial derivatives and set them equal to zero. Thus

F _x = yz - » [ y + 2 z ] = 0

F _y = xz - » [ x + 2 z ] = 0

F _z = xy - » [2 x + 2 y ] = 0

To solve these equations we solve for » in the last equation and substitute it into the first two

and

Multiplying through by 2x + 2y and collecting terms, we obtain

2 y ² z = xy ²

and

2 x ² z = x ² y

These equations can be written in factored form as

y ² ( x - 2 z ) = 0

and

x ² ( y - 2 z ) = 0

From the first equation, we obtain y = 0 and x = 2z. From the second equation we obtain x = 0 and y = 2z. The possible solutions then consist of the pairs

(y = 0, x = 0), (y = 0, y = 2 z ), ( x = 2 z , x = 0), ( x = 2 z , y = 2 z )

These are equivalent to

(y = 0, x = 0), (y = 0, z = 0), ( x = 0, z = 0), ( x = 2 z , y = 2 z )

We still have the restriction equation to be satisfied, which is

xy + 2 xz + 2 yz = 24

The first of the four pairs of partial solutions does not satisfy the restriction equation; hence, it may be discarded. The second and third pairs also fail to satisfy the restriction equation, and they may also be discarded. Hence, we are left with the fourth pair only. Substituting those values into the restriction equation, we obtain

4z ² + 4z ² + 4z ² = 24

This gives z ² = 2 and z = ± . There are therefore two legitimate critical points:

The values of f(x,y,z) = xyz at these two points are 8 and -8 respectively. Since we are interested in maximizing f we are left with the point x = 2 , y = 2 , z = .

This problem grew out of maximizing the volume of a box subject to the restriction that it could not use more than 24 square feet of material. We know from practical considerations that a finite maximum exists; therefore, this solution must produce an absolute maximum.